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Can someone help me with this: Demonstrate by recurrence that: $$ \frac{1}{n!} \leq \left(\frac{e}{n}\right)^n \ \ \ (\forall n>0)$$ Thank you.

rlartiga
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user505452
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  • Erm. What have you tried already? Clearly, you should at least be able to verify the base case for the induction, and attempt to prove the induction step... – Clement C. Dec 06 '17 at 23:39
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    I recommend refactoring this to $\frac{n^n}{n!} \le e^n$. – Chris Dec 06 '17 at 23:45
  • @ClementC. I did verify the case for n=1, after that I multiplied both sides by (1/n+1) so that I get ((1/(n+1)!) But now I'm lost – user505452 Dec 06 '17 at 23:48
  • @Chris Ooh yeah its way simpler that way when I multiply both sides by n/(n+1) which is clearly smaller than e and I get my results. Thank you very much sir! – user505452 Dec 07 '17 at 00:09

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Variant not using induction, since $\ln$ is $\nearrow$ on $\mathbb R^+$ then

$\displaystyle \ln(n!)=\sum\limits_{k=1}^n\ln(k)=\sum\limits_{k=2}^n\ln(k)\ge \int_1^n\ln(t)\mathop{dt}=\bigg[t\ln(t)-t\bigg]_1^n=n\ln(n)-n+1$

Thus $n!\ge n^ne^{-n}e$

Proving by recurrence, we need the intermediate result that $\left(1+\frac 1n\right)^n\le e$, in fact this sequence is increasing and has limit $e$, this is how we get the inequality.

I have to show $(1+\frac1n)^n$ is monotonically increasing sequence

Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $

Now the induction:

$P(1): \dfrac{1^1}{1!}=1\le e\qquad\checkmark$

$P(n+1): \dfrac{(n+1)^{n+1}}{(n+1)!}=\dfrac{(n+1)^n}{n!}\le e^n\dfrac{(n+1)^n}{n^n}\le e^n\times (1+\frac 1n)^n\le e^n\times e=e^{n+1}\qquad\checkmark$

zwim
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